Problem: Multiply the following rational expressions and simplify the result. $\dfrac{x^2-5nx+4n^2}{-3n+3x} \cdot \dfrac{nx^2+16n^3}{12n^3}=$
Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $x^2-5nx+4n^2$, of the first expression can be factored as $(x-4n)(x-n)$ using the sum-product pattern. The denominator, $-3n+3x$, of the first expression can be factored as $3(-n+x)$ by factoring out $3$. The numerator, $nx^2+16n^3$, of the second expression can be factored as $n(x^2+16n^2)$ by factoring out an $n$. The denominator, $12n^3$, of the second expression cannot be factored further. Now the product looks as follows: $\dfrac{(x-4n)(x-n)}{3(-n+x)} \cdot \dfrac{n(x^2+16n^2)}{12n^3}$ To multiply two rational expressions, we multiply across, then simplify: [What's that?] $\begin{aligned} &\phantom{=} \dfrac{(x-4n)(x-n)}{3(-n+x)} \cdot \dfrac{n(x^2+16n^2)}{12n^3} \\\\\\ &= \dfrac{(x-4n)(x-n) \cdot n(x^2+16n^2)}{3(-n+x) \cdot 12n^3} &\text{Multiply across.}\\\\\\\\ &= \dfrac{(x-4n){\cancel{(x-n)}} {\cancel{n}}(x^2+16n^2)}{3{\cancel{(-n+x)}} 12{\cancel{n}} \cdot n^2} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{(x-4n)(x^2+16n^2)}{36n^2} \end{aligned}$ Therefore, the simplified form of the product is $\dfrac{(x-4n)(x^2+16n^2)}{36n^2}$, which is equivalent to $\dfrac{x^3-4nx^2+16n^2x-64n^3}{36n^2}$.